西南交通大学第十三届ACM决赛 E.Shortest Path【思维+Dfs】

题目描述

Today HH becomes a designer, and he faces a problem so he asks you for help.

Treeisland is a country with n cities and n−1 two-way road and from any city you can go to any other cities.

HH the designer is going to design a plan to divide n city into n/2 pairs so that the sum of the length between the n/2 pairs city is minimum.

Now HH has finished it but he doesn't know whether it's true so he ask you to calculate it together.

It's guaranteed that n is even.

输入描述:

 The first line contains an positive integer T(1≤T≤100), represents there are T test cases. 
 For each test case: The first line contains an positive integer n(1≤n≤104), represents the number of cities in Treeisland, it's guarantee that n is even. 
 Then n−1 lines followed. Each line contains three positive integer u, v and len, (u≠v,1≤u≤n,1≤v≤n,1≤len≤109)indicating there is a road of length len between u and v. 
 It's guarantee you can get to any city from any city.

输出描述:

For each test case, output in one line an integer, represent the minimum sum of length.

示例1

输入

241 2 52 3 83 4 661 3 53 2 34 5 44 3 94 6 10

输出

1131

说明

In the first example, you can divide them into (1,2), and (3,4), then the minimum sum of length is 5+6=11
In the second example, you can divide them into (1,3),(2,4),(5,6), hen the minimum sum of length is 5+(3+9)+(10+4)=31
  

题目大意：

给出N个点，让我们将其分成n/2对，每对点的贡献值为两点距离，求最小距离和。

思路：

①我们可以YY一下，对于很多种分配方案，我们可以很容易找到一种情况，就是一条边要么对答案有贡献，要么就是没贡献（也就是求两点距离的时候不会走过这样一条边）。

②题目又是在一棵树上做文章，那么我们肯定要想O（n）的复杂度去Dfs怎么做一做能够得到解。

对于一个点u来讲，如果以这个点u作为根的子树的点的个数为奇数的话，我们知道其子树中肯定有一个点，要和外边一个点相配对才行，那么对应从父亲到当前节点u这条边就一定会有贡献。

③那么过程维护一下，对应到一个点的时候，去判定子树的size是否为奇数即可。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;#define ll long long intstruct node{    int from;    int to;    int w;    int next;}e[350000];int cont;int head[150000];int size[150000];ll output;void add(int from,int to,int w){    e[cont].to=to;    e[cont].w=w;    e[cont].next=head[from];    head[from]=cont++;}void Dfs(int u,int from,int prew){    size[u]=1;    for(int i=head[u];i!=-1;i=e[i].next)    {        int v=e[i].to;        int w=e[i].w;        if(v==from)continue;        Dfs(v,u,w);        size[u]+=size[v];    }    if(size[u]%2==1)output+=prew;}int main(){    int n;    int t;scanf("%d",&t);    while(t--)    {        output=0,cont=0;        scanf("%d",&n);        memset(head,-1,sizeof(head));        for(int i=1;i<=n-1;i++)        {            int x,y,w;scanf("%d%d%d",&x,&y,&w);            add(x,y,w);add(y,x,w);        }        Dfs(1,-1,0);        printf("%lld\n",output);    }}